Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are … - Vidyadip

Question

Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.

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Answer

Vertices A, B, C of a triangle are A (1, 2, 7), B(2, 6, 3) and C(3, 10, -1) respectively.
$\therefore\ \ \ \text{Position vector of point}\ \text{A}=\overrightarrow{\text{OA}}$ $=(1, 2, 7)=\hat{i}+2\hat{j}+7\hat{k}$
$\text{Position vector of point}\ \text{B}=\overrightarrow{\text{OB}}$ $=(2, 6, 3)=2\hat{i}+6\hat{j}+3\hat{k}$

$\text{Position vector of point}\ \text{C}=\overrightarrow{\text{OC}}$ $=(3, 10, -1)=3\hat{i}+10\hat{j}-\hat{k}$
Now $\ \ \overrightarrow{\text{AB}}$ = Position vector of point B - Position vector of point A

$=2\hat{i}+6\hat{j}+3\hat{k}-\big(\hat{i}+2\hat{j}+7\hat{k}\big)$

$=2\hat{i}+6\hat{j}+3\hat{k}-\hat{i}-2\hat{j}-7\hat{k}=\hat{i}+4\hat{j}-4\hat{k}\ \ \ \ \ \ \text{ ........(i)}$
And $\ \overrightarrow{\text{AC}}$ = Position vector of Point C - Position vector of point A

$=3\hat{i}+10\hat{j}-\hat{k}-(\hat{i}+2\hat{j}+7\hat{k})$

$=3\hat{i}+10\hat{j}-\hat{k}- \hat{i}-2\hat{j}-7\hat{k}$ $=2\hat{i}+8\hat{j}-8\hat{k}=2(\hat{i}+4\hat{j}-4\hat{k})\ \ \ \ \ ......\text{(ii)}$
$\Rightarrow\ \ \overrightarrow{\text{AC}}=2.\overrightarrow{\text{AB}}\ \ \ \big[\text{Using eq. (i)}\big]$
$\Rightarrow\ \text{Vectors}\ \overrightarrow{\text{AB}}\ \text{and}\ \overrightarrow{\text{AC}}\ \text{are collinear and parallel.}$ $\ \ \Big[\because\ \vec{a}=m\vec{b}\Big]$
Thus, Points A, B and C are collinear.
And also vectors $\overrightarrow{\text{AB}}\ \text{and}\ \overrightarrow{\text{AC}}$ have a common point A and hence can't be parallel.

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